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\title{Examples of the direct images of D-modules}
\author{LQW}
%\date{September 27, 2025}

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\begin{document}

\maketitle

\abstract{We study some examples of direct images of $D$-modules.}

\section{Example 1}

\subsection{Statement of the problem}

Let $K$ be a field. %for example $K=\mathbb{C}$. 
Let $A_1=A_{1,x}=K[x,\partial_x]$ be the Weyl algebra on the affine line $K^1_x$ with coordinate $x$. 
Let $F:K^1_x\to K^1_y$ be a polynomial map, i.e., $y=F(x)$ is a polynomial. 
%
Let $P\in A_{1}$ be a linear differential operator. 
Let $M=A_{1}/PA_{1}$ be the right $A_{1}$-module. 
The problem is to find the direct image $F_*M$ in the category of $D$-modules. 

\subsection{Solution preparation - but wrong.}

By definition of direct image of right $D$-modules on an affine space, 
\begin{equation}
\begin{aligned}
F_*(M) &= M\otimes_{A_1}D_{K^1_x\to K^1_y} \\ 
    &= M\otimes_{A_1} K[x]\otimes_{K[y]} K[y,\partial_y] \\ 
    &= (K[x,\partial_x]/PK[x,\partial_x]) \otimes_{A_1} k[x]\otimes_{K[y]} K[y,\partial_y] \\ 
    &\cong K[x]/P(K[x]) \otimes_{K[y]} K[y,\partial_y]. 
\end{aligned}
\end{equation}
Here the isomorphism is given by the right exactness of the tensor operation. 
From the exact sequence 
\begin{equation}
0\to PA_1 \to A_1 \to A_1/PA_1 \to 0
\end{equation}
we get the exact sequence 
\begin{equation}
PA_1\otimes_{A_1}K[x] \to A_1\otimes_{A_1}K[x] \to (A_1/PA_1)\otimes_{A_1}K[x] \to 0.
\end{equation}

The left $A_1$-module $K[x]$ is given by $K[x]=A_1/A_1\partial_x$. 
Thus the left action of $P$ on $f(x)\in K[x]$ is given by $P(f(x))$. 
For example, assume $P=a_2(x)\partial_x^2+a_1(x)\partial_x+a_0(x)$. 
Then in $A_1$, we have 
\begin{equation}
\begin{aligned}
Pf(x) &= \left[a_2(x)\partial_x^2+a_1(x)\partial_x+a_0(x) \right] f(x) \\
&= a_2(x)\partial_x^2 f(x) +a_1(x)\partial_x f(x) +a_0(x) f(x) \\
&= a_2(x)\partial_x [f'(x)+f(x)\partial_x] + a_1(x)[f'(x) + f(x)\partial_x] +a_0(x) f(x) \\
&= a_2(x)[f''(x) + 2f'(x)\partial_x + f(x)\partial_x^2] + a_1(x)[f'(x) + f(x)\partial_x] +a_0(x) f(x) \\
&= a_2(x)f''(x) +a_1(x)f'(x)+a_0(x) + Q\partial_x \\ 
&= P(f(x)) + Q\partial_x, 
\end{aligned}
\end{equation}
where $Q\in A_1$. This says that $Pf(x)\equiv P(f(x)) \,\text{mod}\, A_1\partial_x$. 

\subsection{Solution main step - wrong}

Now we want to explain the structure of the right $A_{1,y}$-module $(K[x]/P(K[x]))[\partial_y]$. 
Note that $K[x]/P(K[x])$ is the set of equivalent classes in $K[x]$, where two polynomials $g_1(x)$ and $g_2(x)$ are equivalent if there exists a polynomial $u(x)$ such that $P(u(x))=g_1(x)-g_2(x)$.  
There is an exact sequence 
\begin{equation}
0\to Ker \to K[x] \overset{P(\cdot)}{\to} K[x] \to Coker \to 0,
\end{equation}
and the cokernel is the object $K[x]/P(K[x])$ we want to study. 
Note that the kernel and cokernel are only $K$-vector spaces. 
We want to see that both are finite dimensional. 

We want to see that the cokernel is a right $K[y]$-module. 
Since the morphism $F:K^1_x\to K^1_y$ is given by $y=F(x)$, the right action of $y$ on $K[x]$ is given by $g(x)y = g(x)F(x)$. 

Note that $K[x]=A_{1,x}/A_{1,x}\partial_x$ is a left $A_1$-module. 
It does not have a right $A_1$-module structure. 
It does not have a left $A_1$ and right $K[x]$ bimodule structure either. Then how are we going to give a right $K[y]$ module and further a right $K[y,\partial_y]$ module structre on it?

Back to the beginning, $M=A_1/PA_1$ is a right $A_1$-module. 
The direct image $$F_*M = M\otimes_{A_1} K[x]\otimes_{K[y]} K[y,\partial_y]$$ is a right $K[y,\partial_y]$-module, which is inherited from the right $A_{1,y}$-module structure of the transfer bimodule. 
Now we want to see more details of this right module structure. 
Especially, can we simplify this direct image module to the form $A_{1,y}/QA_{1,y}$, where $Q\in A_{1,y}$ represents some linear differential operator on the affine space $K^1_y$? 

AI says that we need Weil restriction and Gröbner basis to go further. 

\subsection{Transfer module}

Let $X:=K^1_x\to K^1_y=:Y$ be a polynomial map given by $y=F(x)$. 
The transfer module $D_{X\to Y}$ is a $D_X-D_Y$ bimodule. 
In our example, $D_{X\to Y} = K[x]\otimes_{K[y]} K[y,\partial_y]$. 
We want to make explicit the bimodule structure on this transfer module. 

For example, consider the element $x\otimes y\partial_y$ in $D_{X\to Y}$. 
We want to verify the equality 
\begin{equation}
[\partial_x \cdot (x\otimes y\partial_y) ] \cdot \partial_y 
= \partial_x \cdot [ (x\otimes y\partial_y) \cdot \partial_y ]. 
\end{equation}
%
The left hand side is 
\begin{equation}
\begin{aligned}
\relax [\partial_x \cdot (x\otimes y\partial_y) ] \cdot \partial_y  
&= [ \partial_x(x)\otimes y\partial_y + x\frac{dy}{dx}\otimes \partial_y(y)\partial_y ] \cdot \partial_y \\ 
&= [ 1\otimes y\partial_y + xF'(x)\otimes \partial_y ] \cdot \partial_y \\
&= [ F(x) + xF'(x) ]\otimes \partial_y^2. 
\end{aligned}    
\end{equation}
%
The right hand side is 
\begin{equation}
\begin{aligned}
\partial_x \cdot [(x\otimes y\partial_y) \cdot \partial_y ]
&= \partial_x \cdot (x\otimes y\partial_y^2) \\ 
&= \partial_x(x) \otimes y\partial_y^2 + x\frac{dy}{dx}\otimes \partial_y(y)\partial_y^2 \\ 
&= 1\otimes y\partial_y^2 + xF'(x)\otimes \partial_y^2 \\ 
&= [F(x) + xF'(x)]\otimes \partial_y^2. 
\end{aligned}    
\end{equation}
%
The rule of thumb here is that, when we pass something throught the tensor operation $\otimes$, from right to left we change $y$ into $F(x)$, and from left to right we change $\partial_x$ into $F'(x)\otimes\partial_y$. 


\section{Example 2}
Let $X:=K^1_x\to K^1_y=:Y$ be a polynomial map given by $y=F(x)$. 
Let $P = a_2(x)\partial_x^2+a_1(x)\partial_x+a_0(x)$ be an element in the Weyl algebra $A_{1,x}=K[x,\partial_x]$. In the spirit of change of variables, will we arrive at some element $Q = b_2(y)\partial_y^2+b_1(y)\partial_y+b_0(y)$ in $A_{1,y}=K[y,\partial_y]$? 


\begin{thebibliography}{99}
\bibitem{coutinho} S. C. Coutinho. A primer of algebraic D-modules. Cambridge University Press. 1995. 
\bibitem{sabbah} Claude Sabbah. Introduction to algebraic theory of linear systems of differential equations. Eléments de la théorie des systèmes différentiels, 
Les cours du CIMPA, Travaux en cours vol. 45, Hermann, Paris, 1993.
\end{thebibliography}

\end{document}